Maximum Area of Triangle Formed by Parabola Tangent and Line

To find the required quantities, we first express the area of triangle $EFG$ in terms of the coordinate $y_0$.

1. Coordinates of Points:

• The point $E$ is given on the $y$-axis as: $E = (0, 3)$

• The point $F(x_0, y_0)$ lies on the parabola $y^2 = 16x$ with $0 \le y \le 6$. Therefore: $x_0 = \frac{y_0^2}{16}$ Hence, $F = \left(\frac{y_0^2}{16}, y_0\right)$.

• The equation of the tangent line to the parabola $y^2 = 16x$ at the point $F(x_0, y_0)$ is: $y y_0 = 8(x + x_0)$

• The tangent line intersects the $y$-axis at $G(0, y_1)$. Setting $x = 0$: $y_1 y_0 = 8x_0 = 8\left(\frac{y_0^2}{16}\right) = \frac{y_0^2}{2}$ Since $y_0 \neq 0$ (as the line $y = mx + 3$ cannot intersect the origin $(0,0)$), we divide by $y_0$: $y_1 = \frac{y_0}{2}$ Thus, $G = \left(0, \frac{y_0}{2}\right)$.

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2. Area of $\Delta EFG$: Since both points $E(0, 3)$ and $G\left(0, \frac{y_0}{2}\right)$ lie on the $y$-axis, the segment $EG$ can be taken as the base of the triangle: $\text{Base } EG = \left|3 - \frac{y_0}{2}\right|$

The altitude of the triangle corresponding to this base is the perpendicular distance of $F$ from the $y$-axis, which is its $x$-coordinate: $\text{Height} = x_0 = \frac{y_0^2}{16}$

Thus, the area of $\Delta EFG$ is: $\text{Area } A(y_0) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \left|3 - \frac{y_0}{2}\right| \left(\frac{y_0^2}{16}\right) = \frac{1}{64} |6 - y_0| y_0^2$

Given that $0 \le y_0 \le 6$, we have $6 - y_0 \ge 0$, which simplifies the area function to: $A(y_0) = \frac{1}{64} (6y_0^2 - y_0^3)$

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3. Maximizing the Area: To find the local maximum, we differentiate $A(y_0)$ with respect to $y_0$: $\frac{dA}{dy_0} = \frac{1}{64} (12y_0 - 3y_0^2) = \frac{3y_0}{64}(4 - y_0)$

Setting $\frac{dA}{dy_0} = 0$: $y_0 = 0 \quad \text{or} \quad y_0 = 4$

By the first derivative test:

• For $y_0 < 4$, $\frac{dA}{dy_0} > 0$
• For $y_0 > 4$, $\frac{dA}{dy_0} < 0$

Thus, $A(y_0)$ attains a local maximum at: $y_0 = 4$

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4. Determining the Values:

• Value of $y_0$ (R): $y_0 = 4 \implies \text{R} \rightarrow 2$

• Value of $y_1$ (S): $y_1 = \frac{y_0}{2} = \frac{4}{2} = 2 \implies \text{S} \rightarrow 3$

• Coordinates of $F$: $x_0 = \frac{4^2}{16} = 1 \implies F = (1, 4)$

• Slope $m$ of the line $L$ (P): The line $L$ passes through $E(0, 3)$ and $F(1, 4)$: $m = \frac{4 - 3}{1 - 0} = 1 \implies \text{P} \rightarrow 4$

• Maximum Area of $\Delta EFG$ (Q): $A_{\max} = \frac{1}{64} (6 - 4)(4^2) = \frac{1}{64} \times 2 \times 16 = \frac{32}{64} = \frac{1}{2} \implies \text{Q} \rightarrow 1$

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Conclusion:

• $\text{P} \rightarrow 4$
• $\text{Q} \rightarrow 1$
• $\text{R} \rightarrow 2$
• $\text{S} \rightarrow 3$

This matches option A.